a) The degree of L over K is finite.

b) There exists a nonzero element $a$ of A such that K is equal to $A[a^{-1}]$.

Let S be a generating subset of the A-algebra L. We use induction on the cardinal of S.

First, suppose that the elements of S are all algebraic over K. The degree of L over K is then finite (V, §3, No. 2, p. 18, Theorem 2). Let $(e_i)_{i\in I}$ be a basis of L over K. There exists a nonzero element $a$ of A such that the coordinates with respect to this basis of the elements of S, of the element 1, and of the elements $e_ie_j$ for $i,j$ in I belong to $A[a^{-1}]$. The set of linear combinations $\sum_{i\in I}a_ie_i$ with $a_i\in A[a^{-1}]$ for every $i$ is then a subring of L. It contains A and S by construction and is therefore equal to L. In particular, it contains $Ke_1$, so K is equal to $A[a^{-1}]$.

Now suppose that an element $s$ of S is transcendental over K. Denote by E the field of fractions of the ring $A[s]$. The $A[s$]-algebra L is generated by S $-\{s\}$. By the induction hypothesis, there exists a nonzero polynomial $P\in A[X]$ such that E is equal to $A[s][P(s)^{-1}]$. Let $\overline{K}$ be an algebraic closure of K (V, §4, No. 3, p. 23, Theorem 2). Since the field $\overline{K}$ is infinite (V, §4, No. 1, p. 20, Proposition 3), there exists an element $x$ of $\overline{K}$ such that $P(x)\not= 0$. Let $\varphi : E\rightarrow K$ be the unique homomorphism from the K-algebra $E = A[s][P(s)^{-1}]$ to the K-algebra $\overline{K}$ that sends $s$ to $x$. This is absurd because E is a transcendental extension of K and $\overline{K}$ is an algebraic extension of K. This completes the proof of the theorem.

a) The degree of $A/\mathfrak{m}$ over K is finite.

b) Let Ω be an algebraically closed extension of K. There exists a K-algebra homomorphism from A to Ω with kernel $\mathfrak{m}$.

The K-algebra $A/\mathfrak{m}$ is generated by finitely many elements, and $A/\mathfrak{m}$ is a field. By Theorem 1, the degree of $A/\mathfrak{m}$ is finite; assertion a) follows. Every extension of K of finite degree is isomorphic to a subextension of Ω (V, §4, No. 1, p. 20, Theorem 1); this gives b).

(i) The polynomials $P_i$ do not have a common zero in $\Omega^n$.

(ii) There exists a family $(Q_i)_{i\in I}$, with finite support, of elements of $K[X_1, . . . ,X_n]$ such that $\sum_{i\in I}P_iQ_i= 1$.

Denote by A the ring $K[X_1, . . . ,X_n]$ and by $\mathfrak{a}$ the ideal generated by the polynomial $P_i$. Condition (i) means that there are no K-algebra homomorphisms from $A/\mathfrak{a}$ to Ω. If it is satisfied, then by Corollary 1, the ring $A/\mathfrak{a}$ has no maximal ideals and is therefore zero, and 1 belongs to $\mathfrak{a}$. This proves that (i) implies (ii). The implication (ii) $\Rightarrow$ (i) is clear.